Question: What is the average rate of change of the function $f(x)=\tan(x)$ over the interval $1\leq x \leq t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\tan(t-1)}{t}$ (Choice B) B $\dfrac{\tan(t-1)}{t-1}$ (Choice C) C $\dfrac{\tan(t)-\tan(1)}{t-1}$ (Choice D) D $\dfrac{\tan(t)-\tan(1)}{1}$
This is the formula for the average rate of change of a function $f$ over the interval $[a,b]$ : $\dfrac{f(b)-f(a)}{b-a}$ We are interested in the average rate of change of $f(x)=\tan(x)$ over the interval $1\leq x \leq t$ : $\begin{aligned} &\phantom{=}\dfrac{f(t)-f(1)}{(t)-(1)} \\\\ &=\dfrac{\tan(t)-\tan(1)}{t-1} \end{aligned}$ The average rate of change of the function is $\dfrac{\tan(t)-\tan(1)}{t-1}$. Notice that the average rate of change is calculated just like the slope of the secant line that intersects the graph of the function at the interval's endpoints.